Thread: find solutions to the equation

1. find solutions to the equation

Hi sorry am i doing this right?
find all solutions to the equation

$\displaystyle x^6 + 2x^3 + 1 = 0$
let $\displaystyle u = x^3$
then $\displaystyle u^2 + 2u + 1 = 0$
$\displaystyle = (u+1)^2 = 0$
thus $\displaystyle u^3 = -1$
$\displaystyle = \sqrt{1} e^{\pi + 2\pi m}$

and then i solve for $\displaystyle m = 0..6$
is that correct?

2. Originally Posted by Dgphru
Hi sorry am i doing this right?
find all solutions to the equation

$\displaystyle x^6 + 2x^3 + 1 = 0$
let $\displaystyle u = x^3$
then $\displaystyle u^2 + 2u + 1 = 0$
$\displaystyle = (u+1)^2 = 0$
thus $\displaystyle u^3 = -1$
$\displaystyle = \sqrt{1} e^{\pi + 2\pi m}$

and then i solve for $\displaystyle m = 0..6$
is that correct?
what is this for?
$\displaystyle = \sqrt{1} e^{\pi + 2\pi m}$

3. Originally Posted by bigwave
what is this for?
$\displaystyle = \sqrt{1} e^{\pi + 2\pi m}$
hmm i thought that that's how to solve it.. like to find the roots of the equation? like to put in into polar form? .. but honestly i have no idea what im doing

4. Originally Posted by Dgphru
Hi sorry am i doing this right?
find all solutions to the equation

$\displaystyle x^6 + 2x^3 + 1 = 0$
let $\displaystyle u = x^3$
then $\displaystyle u^2 + 2u + 1 = 0$
$\displaystyle = (u+1)^2 = 0$
thus $\displaystyle u^3 = -1$
$\displaystyle = \sqrt{1} e^{\pi + 2\pi m}$

and then i solve for $\displaystyle m = 0..6$
is that correct?
$\displaystyle -1=e^{\pi(2m+1){i}}$

CB

5. Originally Posted by CaptainBlack
$\displaystyle -1=e^{\pi(2m+1){i}}$

CB
thanks!!